# What does "conformally coupled scalar" mean?

"Conformally coupled scalar $$\phi$$" - I encounter it a lot, but I can't find what it means.

A minimally coupled free scalar field is described by the action $$S[g,\phi] = \frac{1}{2} \int d^D x \sqrt{g} g^{ab} \partial_a \phi \partial_b \phi .$$ However, this theory is not conformally invariant (i.e. invariant under Weyl transformations). In particular, if I rescale $$g \to \Omega^2 g$$ (where $$\Omega$$ is a function), then the action transforms as $$S[\Omega^2 g,\phi] = \frac{1}{2} \int d^D x \Omega^{D-2} \sqrt{g}g^{ab} \partial_a \phi \partial_b \phi .$$ where the $$\Omega^D$$ factor comes from $$\sqrt{g}$$ and $$\Omega^{-2}$$ comes from $$g^{ab}$$. Clearly, the action is not conformally invariant (unless $$D=2$$). One way to fix this is to rescale $$\phi$$ as well by $$\phi \to \Omega^{-\frac{1}{2}(D-2)} \phi$$. We then find $$S[\Omega^2 g,\Omega^{-\frac{1}{2}(D-2)} \phi] = \frac{1}{2} \int d^D x \sqrt{g} [ g^{ab} \partial_a \phi \partial_b \phi + {\cal O} \left( \partial \Omega \right) ] .$$ As is clear, the action is still not invariant under Weyl transformations because of terms proportional to the derivative of $$\Omega$$. To fix this, we add an extra term to the original action $$S[g,\phi] = \frac{1}{2} \int d^D x \sqrt{g} \left( g^{ab} \partial_a \phi \partial_b \phi + \xi R[g] \phi^2 \right) . \tag{1}$$ It is then easy to show that for an appropriate choice of the constant $$\xi$$, this action is conformally invariant. A scalar field which couples to the background metric $$g$$ according to (1) is known as a "conformally coupled scalar" for obvious reasons.

Note that the action for a conformally coupled scalar reduces to the action of a minimally coupled scalar in flat spacetime. However, the two theories have different stress tensors.

1. Show that the action (1) is conformally invariant iff. $$\xi = \frac{D-2}{4(D-1)}$$.